Let f:[0,1] -> [0,1] be defined as follows:
f(x)=1/n if x=m/n where n,m are integers, n is not zero, and m/n is irreducible.
f(x)= 0 if x is irrational.
Prove that f(x) is Riemann integrable on [0,1]

please help me on this.

Jan 2nd 2009, 05:53 AM

HallsofIvy

Quote:

Originally Posted by Kat-M

Let f:[0,1] -> [0,1] be defined as follows:
f(x)=1/n if x=m/n where n,m are integers, n is not zero, and m/n is irreducible.
f(x)= 0 if x is irrational.
Prove that f(x) is Riemann integrable on [0,1]

please help me on this.

What do you have to use? That function, the "modified Dirchlet function" (the Dirichlet function itself is 0 for x rational, 1 for x irrational), can be proved to be continuous exactly on the irrational numbers and 0. That means that it is a bounded function whose set of discontinuities has measure 0. There is a theorem that says such a function is Riemann integrable.

Jan 5th 2009, 06:35 PM

Kat-M

still having a problem

Quote:

Originally Posted by HallsofIvy

What do you have to use? That function, the "modified Dirchlet function" (the Dirichlet function itself is 0 for x rational, 1 for x irrational), can be proved to be continuous exactly on the irrational numbers and 0. That means that it is a bounded function whose set of discontinuities has measure 0. There is a theorem that says such a function is Riemann integrable.

thank you very much for the reply. but i am not really getting it. i ve never heard of Dirchelt and i dont understant what it means to say the set of discontinuities has measure 0. is there any other way? if so i would really appriciate it.